Riemann surface
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Example 10.4 (p.346)
In Example 8.9 in Chapter 8 (P.268), we derived the an equation to describe a mass that is attached to a spring that would break when its elongation reached $0.03 \mathrm{~m}$ during resonant vibration of the springmass system. We need to determine the time $t_f$ at which the spring breaks from Equation (a):
$$\left(0.05-\frac{t_f}{20}\right) \cos 10 t_f+\frac{1}{200} \sin 10 t_f-0.03=0$$

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Solution:
We will use the Newton-Raphson’s method to solve the unknown quantity $t_f$ in Equation (a) by first assuming a solution on $\mathrm{t}{\mathrm{f}}=0.75$. We made this assumed solution based on a crude approximated value of $\mathrm{t}{\mathrm{f}}=0.7$ in Example 8.9.
Again, let us replace the unknown quantity $t_f$ in Equation (a) by conventional unknown symbol $\mathrm{x}$ in the following alternative form:
$$\begin{gathered} \left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03=0 \ f(x)=\left(0.05-\frac{x}{20}\right) \cos 10 x+\frac{1}{200} \sin 10 x-0.03 \ f^{\prime}(x)=-(0.05-0.5 x) \sin 10 x \end{gathered}$$
Thus, the estimated root $x_{i+1}$ after the previously estimated root $x_i$ may be computed by using the expression in Equation (10.5), as will be shown in the next slide.

Example 10.9 (p.358)
Evaluate the following integral by using the Gaussian quadrature in Equation (10.14).
$$I=\int_0^\pi \cos x d x$$

We have the function $y(x)=\cos x$ over the integration limits $x_a=0$ and $x_b=\pi$. The transformation of coordinates makes use of the relationship $x=\frac{\pi}{2} \xi+\frac{\pi}{2}$ from Equation (10.13), from which we get: $y(x)=\cos x=F(\xi)=\cos \left(\frac{\pi}{2} \xi+\frac{\pi}{2}\right)=\sin \left(\frac{\pi}{2} \xi\right)$
Also, from Equation (10.14) with the use of the trigonometric relationships such as: $\sin \left(\frac{\pi}{2}+\theta\right)=\cos \theta$ and $\cos \left(\frac{\pi}{2}+\theta\right)=\sin \theta$
We may arrive at the following expression for integrating I in Equation (a) using Gaussian quadrature:
$$I=\int_0^\pi \cos x d x=\int_{-1}^1\left[\sin \left(\frac{\pi}{2} \xi\right)\left(\frac{\pi}{2} d \xi\right)\right]=\frac{\pi}{2} \int_{-1}^1 \sin \frac{\pi}{2} \xi d \xi=\frac{\pi}{2} \sum_{i=1}^n H_i \sin \left(\frac{\pi}{2} a_i\right)$$
Let us take, for example, 3 sampling points, i.e., $\mathrm{n}=3$ from Table 10.3 on P.357 with:
$$\begin{array}{lll} \mathrm{a}_1=0 & \mathrm{a}_2=+0.77459 & \mathrm{a}_3=-0.77459 \ \mathrm{H}_1=0.88888 & \mathrm{H}_2=0.55555 & \mathrm{H}_3=0.55555 \end{array}$$
Substituting the above numbers into Equation (b) will lead to the solution:
\begin{aligned} I & =\frac{\pi}{2}\left[0.88888 \sin (0)+0.55555 \sin \left(\frac{\pi}{2} \times 0.77459\right)+0.55555 \sin \left(-\frac{\pi}{2} \times 0.77459\right)\right] \ & =\frac{\pi}{2}[0.55555 \sin (1.2167)-0.55555 \sin (1.2167)]=0 \end{aligned}
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